Question: Let $a_1,$ $a_2,$ $\dots,$ $a_{12}$ be positive real numbers such that $a_1 + a_2 + \dots + a_{12} = 1.$  Find the minimum value of
\[\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_{12}}.\]
Answer: By Cauchy-Schwarz,
\[(a_1 + a_2 + \dots + a_{12}) \left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_{12}} \right) \ge (1 + 1 + \dots + 1)^2 = 12^2 = 144,\]so
\[\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_{12}} \ge 144.\]Equality occurs when $a_i = \frac{1}{12}$ for all $i,$ so the minimum value is $\boxed{144}.$